# How much kinetic energy does a helicopter use in a hover? [duplicate]

A helicopter just circulates air in a hover and maintains a stable altitude. So, how much energy is used to do this? Using the standard equation $$KE = \frac12 mv^2$$; then the kinetic energy used would be $$0.5$$ times the mass of air displaced down each second ($$m$$), times the vertical velocity of this air squared ($$v^2$$). Is this correct?

I couldn't find this in any engineering textbook. Estimates (guesses) from fuel burn is helpful, but are very inaccurate; as they rely on estimates (assumptions) of how efficient the fuel conversion to mechanical energy is, how efficient the motor is at using this energy, and then how efficient the rotors are at displacing air down. Which is too many assumptions.

## marked as duplicate by G. Smith, Aaron Stevens, Jon Custer, Qmechanic♦Aug 9 at 6:23

• You are neglecting the kinetic energy used to spin the blades, they have rotational kinetic energy – Triatticus Aug 8 at 21:43
• Here is a good place to start: grc.nasa.gov/www/k-12/airplane/propth.html – David White Aug 8 at 21:45
• The amount of energy used to keep a helicopter aloft depends on the size of the main propeller(s). A large helicopter is much more energy efficient than a small quadcopter. Suggest that you look up and read some basic information on the physics of helicopter flight. – Samuel Weir Aug 8 at 21:46
• Possible duplicate of Calculate work done by a hovering helicopter over time – Aaron Stevens Aug 8 at 21:50
• Hey @NicholasLandell-Mills: for smaller vs. heavier, yes. All else being equal, after adjusting disk speed, a helicopter with a large disk diameter should be more efficient than one with a small disk diameter (all else being equal, especially weight). Think sailplane vs. some 4:1 aspect ratio sporty plane. Getting all that "else" stuff to be equal may not be practical, though. – TimWescott Aug 8 at 22:00

Using the standard equation $$KE = \frac12 mv^2$$; then the kinetic energy used would be $$0.5$$ times the mass of air displaced down each second ($$m$$), times the vertical velocity of this air squared ($$v^2$$). Is this correct?
No, unless you consider heat to be a form of kinetic energy. Air is a massively complicated, turbulent fluid, and the air the helicopter blades are interacting with isn’t pushed down as a single monolithic body moving at a fixed speed $$v$$, but instead gets thrown around in all sorts of directions, creates vortices, and much of it ends up dissipating as heat and sound waves, or moving in directions that don’t generate lift. So unless you’re willing to go down to the molecular level and add up the kinetic energy of lots of air molecules bouncing around in all directions (good luck with that), no, your suggested method of calculating the energy required for hovering is not correct.